giải hệ phương trình3 căn 5 X-4Y=15-2 căn 7 và 3X-2Y=-3 01/11/2021 Bởi Arianna giải hệ phương trình3 căn 5 X-4Y=15-2 căn 7 và 3X-2Y=-3
Đáp án: \(\left\{ \begin{array}{l}x = \dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}}\\y = \dfrac{{45 – 6\sqrt 7 + 9\sqrt 5 }}{{2\left( {3\sqrt 5 – 6} \right)}}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\left\{ \begin{array}{l}3\sqrt 5 x – 4y = 15 – 2\sqrt 7 \\3x – 2y = – 3\end{array} \right.\\ \to \left\{ \begin{array}{l}3\sqrt 5 x – 4y = 15 – 2\sqrt 7 \\ – 6x + 4y = 6\end{array} \right.\\ \to \left\{ \begin{array}{l}\left( {3\sqrt 5 – 6} \right)x = 21 – 2\sqrt 7 \\3x – 2y = – 3\end{array} \right.\\ \to \left\{ \begin{array}{l}x = \dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}}\\y = \dfrac{{3x + 3}}{2} = \dfrac{{3.\dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}} + 3}}{2} = \dfrac{{63 – 6\sqrt 7 + 9\sqrt 5 – 18}}{{2\left( {3\sqrt 5 – 6} \right)}}\end{array} \right.\\ \to \left\{ \begin{array}{l}x = \dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}}\\y = \dfrac{{45 – 6\sqrt 7 + 9\sqrt 5 }}{{2\left( {3\sqrt 5 – 6} \right)}}\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left\{ \begin{array}{l}
x = \dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}}\\
y = \dfrac{{45 – 6\sqrt 7 + 9\sqrt 5 }}{{2\left( {3\sqrt 5 – 6} \right)}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
3\sqrt 5 x – 4y = 15 – 2\sqrt 7 \\
3x – 2y = – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3\sqrt 5 x – 4y = 15 – 2\sqrt 7 \\
– 6x + 4y = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {3\sqrt 5 – 6} \right)x = 21 – 2\sqrt 7 \\
3x – 2y = – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}}\\
y = \dfrac{{3x + 3}}{2} = \dfrac{{3.\dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}} + 3}}{2} = \dfrac{{63 – 6\sqrt 7 + 9\sqrt 5 – 18}}{{2\left( {3\sqrt 5 – 6} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{21 – 2\sqrt 7 }}{{3\sqrt 5 – 6}}\\
y = \dfrac{{45 – 6\sqrt 7 + 9\sqrt 5 }}{{2\left( {3\sqrt 5 – 6} \right)}}
\end{array} \right.
\end{array}\)